Question

If $z=x+iy$ and $|z-ai|=|z+ai|$ then the locus of $z$ is

(Given : a is a real number)

• A. $y=0$
• B. $x=0$
• C. $x=y$
• D. $x^2+y^2=1$

Answer 1

The correct option is A $y=0$

Substitucting $z=x+iy$
$\Rightarrow |x+iy-ai|=|x+iy+ai|$
$\Rightarrow x^2+(y-a{)}^2=x^2+(y+a{)}^2$
$\Rightarrow 2ay=0$
$\Rightarrow y=0$